Problem 4117 from Crux Mathematicorum

Problem

Problem 4117 from Crux Mathematicorum

Solution 1

Here's a few useful facts:

  1. $\text{arcsinh}(x)=\ln\left(\sqrt{x^2+1}+x\right);$

  2. $\sqrt{\sinh^2(x)+1}=\cosh(x),$

  3. $\sinh(x)\cosh(y)+\sinh(y)\cosh(x)=\sinh(x+y).$

For any $n\ge 0,$ denote $\text{arcsinh}(x_n)=y_n.$ Then $x_n=\sinh(y_n).$ Further, $y_0=0,$ $y_1=\ln (\sqrt{2}+1)$ and

$\begin{align} \sinh)y_{n+1}&=\sinh(y_n)\cdot\sqrt{\sinh^2(y_{n-1})+1}+\sinh(y_{n-1})\cdot\sqrt{\sinh^2(y_n)+1}\\ &=\sinh(y_n)\cosh (y_{n-1})+\sinh(y_{n-1})\cosh (y_n)\\ &=\sinh(y_n+y_{n-1}), \end{align}$

implying $y_{n+1}=y_{n}+y_{n-1}.$ That's a famous second-degree recurrence, with the general solution in the form

$\displaystyle y_n=\alpha\left(\frac{1+\sqrt{5}}{2}\right)^n+\beta\left(\frac{1-\sqrt{5}}{2}\right)^n,$ $n\ge 0.$

From $y_n=0,$ $\alpha+\beta=0;$ from $y_1=\ln (\sqrt{2}+1),$ $\alpha-\beta=2\ln (\sqrt{2}+1)\cdot\frac{1}{\sqrt{5}},$ so that $\displaystyle \alpha\ln (\sqrt{2}+1)\cdot\frac{1}{\sqrt{5}}$ and $\displaystyle \beta\ln (\sqrt{2}+1)\cdot\frac{1}{\sqrt{5}},$ giving

$\displaystyle y_n=\frac{\ln(\sqrt{2}+1)}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right],$

for all $n\ge 0.$ If $\displaystyle F_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right],$

is the $n^{th}$ Fibonacci number, then $y_n=(\ln (\sqrt{2}+1)F_n,$ from which

$\displaystyle x_n=\sinh[\ln(\sqrt{2}+1)F_n]=\frac{(\sqrt{2}+1)^{F_n}-(\sqrt{2}-1)^{F_n}}{2}.$

Solution 2

We perform the transformation $x_n=\sinh\left( f_n\right)$

$\sinh\left(f_{n+1}\right)=\sinh (f_n) \cosh (f_{n-1})+\sinh (f_{n-1}) \cosh (f_n)$

which reduces to

$\sinh (f_{n+1})=\sinh (f_n+f_{n-1})$

Taking $\text{arcsinh}$ on both sides, we have a Fibonacci sequence: $f_{n+1}=f_n+f_{n-1}$ with solution $\displaystyle f_n=\frac{\varphi ^n-\varphi ^{-n} (\cos (\pi n) )}{\sqrt{5}}$ where $\varphi$ is the golden ratio. Finally, since $x_1=1,$

$\displaystyle x_n=\sinh \left(\text{arcsinh}(1)\frac{\displaystyle \left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)^n-\left(\frac{2}{\sqrt{5}+1}\right)^n \cos (\pi n)}{\sqrt{5}}\right)$

Here's numerical verification of that formula:

Problem 4117 from Crux Mathematicorum, Verification

Solution 3

Let's set $y_n=x_n+\sqrt{x_n^2}.$ Then $\displaystyle \sqrt{x_n^2+1}-x_n=\frac{1}{y_n}.$ Then $\displaystyle x_n=\frac{1}{2}\left(y_n-\frac{1}{y_n}\right)$ and $\displaystyle \sqrt{x_n^2+1}=\frac{1}{2}\left(y_n+\frac{1}{y_n}\right).$

The given equation van be rewritten as $\displaystyle y_{n+1}-\frac{1}{y_{n+1}}=y_ny_{n-1}-\frac{1}{y_ny_{n-1}}$ which is equivalent to $\displaystyle (y_{n+1}-y_{n-1}y_n)\left(1+\frac{1}{y_{n-1}y_n}\right)=0.$

Since necessarily, $y_n\gt 0,$ it follows that $y_{n+1}=y_{n}y_{n-1}.$ Let $z_n=\log y_n.$ Then $z_{n+1}=z_n+z_{n-1},$ $z_0=0,$ $z_1=\log (\sqrt{2}+1),$ so that $\displaystyle z_n=\frac{1}{\sqrt{5}}\left(\varphi^n-(-\varphi)^{-n}\right)\log(1+\sqrt{2}).$ Finally,

$y_n=(1+\sqrt{2})^{\frac{1}{\sqrt{5}}(\varphi^n-(-\varphi)^{-n})}.$

Acknowledgment

This is Problem 4117 from the Canadian Crux Mathematicorum; the problem was posed by Martin Lukarevski. I am grateful to Leo Giugiuc who kindly shared with me the problem and his solution (Solution 1.) Solution 2 is by N. N. Taleb; Solution 3 is by Kunihiko Chikaya.

 

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